Matematika/Išvestinė polinėje koordinačių sistemoje: Skirtumas tarp puslapio versijų

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:<math>\tan\mu=\tan(\phi-\theta)=\frac{\tan\phi-\tan\theta}{1+\tan\phi\tan\theta}=\frac{\frac{\rho'\sin\theta+\rho\cos\theta}{\rho'\cos\theta-\rho\sin\theta}-\frac{\sin\theta}{\cos\theta}}{1+\frac{\rho'\sin\theta+\rho\cos\theta}{\rho'\cos\theta-\rho\sin\theta} \cdot \frac{\sin\theta}{\cos\theta}}=\frac{\frac{(\rho'\sin\theta+\rho\cos\theta)\cos\theta-(\rho'\cos\theta-\rho\sin\theta)\sin\theta}{(\rho'\cos\theta-\rho\sin\theta)\cos\theta}}{\frac{(\rho'\cos\theta-\rho\sin\theta)\cos\theta+(\rho'\sin\theta+\rho\cos\theta)\sin\theta}{(\rho'\cos\theta-\rho\sin\theta)\cos\theta} }=</math>
:<math>=\frac{(\rho'\sin\theta+\rho\cos\theta)\cos\theta-(\rho'\cos\theta-\rho\sin\theta)\sin\theta}{(\rho'\cos\theta-\rho\sin\theta)\cos\theta+(\rho'\sin\theta+\rho\cos\theta)\sin\theta }=\frac{\rho'\sin\theta\cos\theta+\rho\cos^2\theta-\rho'\cos\theta\sin\theta+\rho\sin^2\theta}{(\rho'\cos^2\theta-\rho\sin\theta\cos\theta+\rho'\sin^2\theta+\rho\cos\theta\sin\theta }=</math>
:<math>=\frac{\rho\cos^2\theta+\rho\sin^2\theta}{(\rho'\cos^2\theta+\rho'\sin^2\theta }=\frac{\rho}{\rho'},</math>
:arba
:<math>\rho'\tan\mu=\rho,</math>
:<math>\rho'=\frac{\rho}{\tan\mu}=\rho\cot\mu,</math>
:<math>\rho'_{\theta}=\rho\cot\mu.</math>